Number System for Digital Computers - International Journal ...

INTERNATIONAL JOURNAL OF SCIENTIFIC & TECHNOLOGY RESEARCH VOLUME 9, ISSUE 04, APRIL 2020 ISSN 2277-8616

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Number System for Digital Computers

Dr. Ms. Shabnam S. Mahat, Dr. Mahadev K. Patil

Abstract : A number system is the method or system of representing the digits in the computer system. The digital computer represents data in

binary forms. The total number of digits used in a number system is called its base or radix. The base is written after the number as subscript; for

example

Binary number system (Base 2), like 101102 (10110 base 2).

Octal number system (Base 8), like 768 (76 base 8).

Decimal number system (Base 10), like 56010 (560 to base of 10).

Hexadecimal number system (Base 16), like 5AC16 (5AC to base of 16).

This paper discusses the binary Addition, Subtraction, Multiplication and Division that will be useful for digital computers, and the field of computer

science.

Keywords: Binary Addition, Binary Subtraction, Binary Multiplication and Binary Division.

————————————————————

1 OBJECTIVE 1. To study a basic arithmetic operation of binary number

system.

2. To simplify the calculation for better understanding of

students.

2 BINARY ADDITION Addition is one of the easy operations of the basic

arithmetic operations. The binary addition operation is

similar to the decimal system. Rules of binary addition are

as follows.

Rules of Addition

Number of

Times Addition

Decimal

Equivalent

Binary

Equivalent

Carry

Generated

Addition

value

1 Time 1 1 1 1 0 1

2 Times 1 1+1 2 10 1 0

3 Times 1 1+1+1 3 11 1 1

4 Times 1 1+1+1+1 4 100 10 0

5 Times 1 1+1+1+1+1 5 101 10 1

6 Times 1 1+1+1+1+1+1 6 110 11 0

7 Times 1 1+1+1+1+1+1+1 7 111 11 1

8 Times 1 1+1+1+1+1+1+1+1 8 1000 100 0

9 Times 1 1+1+1+1+1+1+1+1+1 9 1001 100 1

10 Times 1 1+1+1+1+1+1+1+1+1+1 10 1010 101 0

Solved Example of Binary Addition

Q1 ) 111(2) + 111 (2) = ? (2)

1 1 1

+ 1 1

Carry + 1 1 1

Addition 1 0 1 0

Binary No Decimal equivalent

1st Number 111 7

2nd

Number 11 3

Addition 1010 10

Answer is :

Q2 ) 11001(2) + 111 (2) = ? (2)

111 (2) + 11 (2) = 1010 (2)

Carry Generated

eneratedGenerat

ed

Step1: addition


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1 1 0 0 1

+ 1 1 1

Carry 1 1 1 1 1

Addition 1 0 0 0 0 0


1st Number 11001 25

2nd

Number 111 7

Addition 100000 32

Answer is :

Q3 ) 11001(2) + 11100(2) + 1000 (2) = ? (2)

1 1 0 0 1

+ 1 1 1 0 0

+ 1 0 0 0

Carry 1 1

Addition 1 1 1 1 0 1


1st Number 11001 25

2nd

Number 11100 28

3rd

Number 1000 8

Addition 111101 61

Answer is :

Q4 ) 11011(2) + 11110(2) + 1000 (2) = ? (2)

1 1 0 1 1

+ 1 1 1 1 0

+ 1 0 0 0

Carry 10 10 1 1

Addition 10 0 0 0 0 1


1st Number 11011 27

2nd

Number 11110 30

3rd

Number 1000 8

Addition 1000001 65

Answer is :

11001 (2) + 111 (2) = 100000 (2)

Carry Generated

eneratedGenerat

ed

Step1: addition

11001(2) + 11100(2) + 1000 (2)= 111101 (2)

Carry Generated

eneratedGenerat

ed

Step1: addition

11011(2) + 11110(2) + 1000 (2)= 1000001

(2)

Carry Generated

eneratedGenerat

ed

Step1: addition


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Binary Addition Practice Questions

QUESTIONS ANSWERES

1. 11001101(2) + 1110011 (2) = ? (2) 101000000

2. 1010101(2) + 1111111 (2) = ? (2) 11010100

3. 111000(2) + 101101 (2) = ? (2) 1100101

4. 101011(2) + 101100 (2) + 100001 (2) = ? (2) 1111000

5. 111000(2) + 000111 (2) + 101010 (2) = ? (2) 1101001

BINARY SUBTRACTION

Binary subtraction is addition of 1’st compliment of

subtrahend with minuend, first make 1’st compliment of

subtrahend and add with minuend using addition rules, if

carry generated, then ignore this carry FROM OBTAINED

addition and add with itself which is positive(+ve) final

result. And if carry not generated, then make 1’st

compliment of obtained result which is negative (-ve) final

result.

Steps of Binary Subtraction

Step1: 1st compliment of subtrahend add with the minuend

IF Carry generated

Step2: If carry generated, add in the obtained addition.

Step 3: Ignore carry, Final +V Result.

IF Carry not generated

Step2: If carry not generated, make 1’st compliment of

obtained result.

Step 3: Final -V Result.

What is the 1’st compliment?

The conversion of 1 into 0 , and 0 into 1 is called as 1’st compliment

e.g. 10010 1’st compliment is 01101

Solved Example of Binary subtraction

Q1 ) 111(2) - 101 (2) = ? (2)

Let us solve

A= 1 1 1 A as it is

+

1st compliment of B = 0 1 0

carry 1 1

Addition 0 0 1

0 0 1 Addition o A and 1st compliment of B

+ 1 Add Carry obtained after Addition of A

and 1st compliment of B

carry 1 Ignore carry

Step1: 1st compliment of

subtrahend add with the

minuend

Step2: If carry generated,

add in the obtained addition


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Addition +V Result 0 1 0 Final +V Result


1st Number 111 7

2nd

Number 101 5

Subtraction 010 +3

Answer is :

Q2 ) 1101(2) - 1010 (2) = ? (2)

Let us Solve

A= 1 1 0 1 A as it is

+

1st compliment of B = 0 1 1 0

carry 1 1

Addition 0 0 1 1

0 0 1 1 Addition o A and 1st compliment of B

+ 1 Add Carry obtained after Addition of A

and 1st compliment of B

carry 1 1 Ignore carry

Addition +V Result 0 1 0 0 Final +V Result


1st Number 1101 13

2nd

Number 1001 9

Subtraction 0100 +4

Answer is :

Q3 ) 101(2) - 111 (2) = ? (2)

Let us Solve

A= 1 0 1 A as it is

+

1st compliment of B = 0 0 0

carry absent .. 1 0 1

Final - V Result - 0 1 0


1st Number 101 5

2nd

Number 111 7

Subtraction -010 -2

Answer is :

Q4 ) 1001(2) - 1101 (2) = ? (2)

Let us Solve

A= 1 0 0 1 A as it is

111 (2) - 101 (2) = 010 (2)



minuend

Step2: If carry generated,

add in the obtained addition

1101 (2) - 1001 (2) = 0100

(2)



minuend

Step2: If carry not generated,

1st compliment of result

101 (2) - 111 (2) = - 010 (2)



minuend


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+

1st compliment of B = 0 0 1 0

carry absent -- 1 0 1 1

Final - V Result 0 1 0 0


1st Number 101 5

2nd

Number 111 7

Subtraction -010 -2

Answer is :

Binary Subtraction Practice Questions

QUESTIONS ANSWERES

1. 110010(2) - 11001 (2) = ? (2) + 11001

2. 10100(2) - 110010 (2) = ? (2) - 11110

3. 1111(2) - 11001 (2) = ? (2) -1010

4. 1100 (2) - 1001 (2) = ? (2) - 0011

5. 100000(2) - 101000 (2) = ? (2) -1000

6. 1010000(2) - 101000 (2) = ? (2) + 101000

7. 110000(2) - 110011 (2) = ? (2) +11

8. 10010(2) - 100000 (2) = ? (2) -1110

9. 10100(2) - 110000 (2) = ? (2) -11100

BINARY MUTIPLICATION

Binary Multiplication is same as binary addition. The

multiplier contains only 0s and 1s, so each multiplication

step produces either zeros or a copy of the multiplicand and

finally performs addition of all steps produces after

multiplication.

Solved Example of Binary Multiplication

Q1 ) 1100(2) × 101 (2) = ? (2)

1 1 0 0

× 1 0 1

1 1 0 0 Step 1 : 1 MULTIPLY WITH 1100 ,THUS AS IT IS 1100

+ 0 0 0 0 * Step 2 : 1 place remain empty i.e.(*), 0 MULTIPLY WITH

1100 , Thus all numbers will be 0000

+ 1 1 0 0 * * Step 3 : 2 place remain empty i.e.(*), 1 MULTIPLY WITH


1 1 1 1 0 0 Step 4 : make addition of the number which get from Step

1,2 and 3


Multiplicand 1100 12

Multiplayer 101 5

Addition 111100 60

Step1 : 1100 × 1 =1100 Step2: 1100 × 0 =0000

Step3: 1100 × 1

=1100 Step4: Make addition

Step2: If carry not generated,

1st compliment of result

101 (2) - 111 (2) = - 010 (2)


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Answer is :

Q2 ) 1101(2) × 110 (2) = ? (2)

1 1 0 1

× 1 1 0

0 0 0 0 Step 1 : 0 MULTIPLY WITH 1101 , Thus all numbers will

be 0000



+ 1 1 0 1 * * Step 3 : 2 place remain empty i.e.(*), 1 MULTIPLY WITH


1 1

1 0 0 1 1 1 0 Step 4 : make addition of the number which get from Step

1,2 and 3



Multiplayer 110 6

Addition 1001110 78

Answer is

Q3 ) 1111(2) × 11 (2) = ? (2)

1 1 1 1

× 1 1

1 1 1 1 Step 1 : 1 MULTIPLY WITH 1111 , Thus all numbers will

be 1111



1 1 1 1

1 0 1 1 0 1 Step 3 : make addition of the number which get from Step

1 and 2



Multiplayer 11 3

Addition 111100 60

Answer is :

Q4 ) 111(2) × 111 (2) = ? (2)

1 1 1

× 1 1

1100 (2) × 101 (2) = 111100 (2)

Step2: 111 ×1 =111 Step1 : 111 × 1 =111

Step1 : 1101 × 0 =0000

1101 (2) × 110 (2) = 1001110

(2)

Step2: 1101 × 1 =1101

Step3: 1101 × 1

=1101

Step4: Make addition

Step1 : 1111 × 1 =1111

1101 (2) × 110 (2) = 1001110

(2)

Step2: 1111 × 1 =1111


Carry

Generated

Carry

Generated


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1 1 1 Step 1 : 1 MULTIPLY WITH 111 ,THUS AS IT IS 111

+ 1 1 1 * Step 2 : 1 place remain empty i.e.(*), 1 MULTIPLY WITH


1 1 1

1 0 1 0 1 Step 3 : make addition of the number which get from Step

1,2 and 3


Multiplicand 111 7

Multiplayer 11 3

Addition 10101 21

Answer is :

Binary Multiplication Practice Questions

QUESTIONS ANSWERES

10. 100111(2) × 1011 (2) = ? (2) 110101101

11. 10101(2) × 11 (2) = ? (2) 111111

12. 1100(2) ×100 (2) = ? (2) 110000

13. 1111100(2) × 10 (2) = ? (2) 11111000

14. 11010(2) × 101 (2) = ? (2) 10000010

15. 1011110(2) × 101 (2) = ? (2) 11100110

16. 11110(2) × 111 (2) = ? (2) 10010110

17. 100000(2) × 10 (2) = ? (2) 1000000

18. 101101(2) × 11 (2) = ? (2) 10000111

19. 10110(2) × 1010 (2) = ? (2) 11011100

BINARY DIVISION Division is one of the difficult operations of the basic

arithmetic operations. The binary division operation is

similar to the decimal system, except the base 2 system.

Solved Example of Binary Division

Q1 ) 11000(2) ÷ 101 (2) = ? (2)

Step

1

Step

2

Step

3

1 0 1

1 0 1 1 1 0 0 1 Step 1 : First 3 Digit of dividend i.e.110 (2) =6(10) is

greater than divisor i.e.101(2) =5(10) Thus quotient is 1,

subtract (6-5=1) Binary equivalent of 1 is 1 - 1 0 1

1 0 Step 2 : Take next number (0) from dividend ,10(2)

=2(10) is less than divisor i.e. 101(2) =5(10). Thus quotient

is 0, get the (10) as it is without subtraction - 1 0 1

1 0 1 Step 3 : Take next number (1) from dividend, 101(2)

=5(10) is equal to divisor i.e.101(2) =5(10). Thus quotient

is 1, after subtraction, reminder is 0 - 1 0 1

0 Step 4 : Remainder is 0 . Thus stop.

Divisor

Dividend Quotient

111 (2) × 11 (2) = 10101 (2)

Carry Generated

eneratedGenerat

ed



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Dividend 11001 25

Divisor 101 5

Quotient 101 5

Remainder 0 0

Answer is :

Q2 ) 11000(2) ÷ 101 (2) = ? (2)

Step

1

Step

2

Step

3

0 1 0

1 1 0 1 1 0 0 1 0 0 Step 1 : First 4 Digit of dividend i.e.1001 (2) =9(10)

is less than divisor i.e.1101(2) =13(10) Thus

quotient is 0, get the (1001) as it is without

subtraction

- 1 1 0 1

1 0 0 1 0 Step 2 : Take next number (0) from dividend

,10010(2) =18(10) is greater than divisor i.e. 1101(2)

=13(10). Thus quotient is 1, subtract (18-13=5)

Binary equivalent of 5 is 101(2)

- 1 1 0 1

0 1 0 1 0 Step 3 : Take next number (0) from dividend,

1010(2) =10(10) is less than to divisor i.e.1101(2)

=13(10). Thus quotient is 0, get the (1010) as it is

without subtraction

- 1 1 0 1

1 0 1 0 Step 4 : Remainder 1010(2) =10(10) is less than

divisor i.e. 1101(2) =13(10). Thus stop.


Dividend 100100 36

Divisor 1101 13

Quotient 010 2

Remainder 1010 10

Answer is :

Q3 ) 111011 (2) ÷ 11 (2) = ? (2)

Step

1

Step

2

Step

3

Step

4

Step

5

1 0 0 1 1

1 1 1 1 1 0 1 1 Step 1 : First 2 Digit of dividend i.e.11 (2)

=3(10) equal to divisor i.e.11(2) =3(10). Thus

quotient is 1, after subtraction, reminder is 0 - 1 1

0 0 1 Step 2 : Take next number (1) from

dividend , 1(2) =1(10) is less than divisor i.e.

11(2) =3(10). Thus quotient is 0, get the (1) as

it is without subtraction

- 1 1

1 0 Step 3 : Take next number (0) from

dividend, 10(2) =2(10) is less than to divisor

i.e.11(2) =3(10). Thus quotient is 0, get the

(10) as it is without subtraction

- 1 1


dividend , 101(2) =5(10) is greater than divisor

i.e. 11(2) =3(10). Thus quotient is 1, subtract

(5-3=2) Binary equivalent of 2 is 10(2)

- 1 1


dividend , 101(2) =5(10) is greater than divisor

i.e. 11(2) =3(10). Thus quotient is 1, subtract

(5-3=2) Binary equivalent of 2 is 10(2)

- 1 1

1 0 Step 6 : Remainder 10(2) =2(10) is less than


11000(2) ÷ 101 (2) = 101 (2)

100100(2) ÷ 1101 (2) = 010 (2)


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Dividend 111011 59

Divisor 11 3

Quotient 10011 19

Remainder 10 2

Answer is

Q4 ) 100001 (2) ÷ 110 (2) = ? (2)

Step

1

Step

2

Step

3

Step

4

0 1 0 1

1 1 0 1 0 0 0 0 1 Step 1 : First 3 Digit of dividend i.e.100 (2)

=4(10) is less than to divisor i.e.110(2) =6(10).

Thus quotient is 0, get the (100) as it is

without subtraction

- 1 1 0

1 0 0 0 Step 2 : Take next number (0) from dividend

,1000(2) =8(10) is greater than divisor i.e. 110(2)



- 1 1 0

1 0 0 Step 3 : Take next number (0) from dividend,

100(2) =4(10)) is less than to divisor i.e.110(2)

=6(10). Thus quotient is 0, get the (100) as it is

without subtraction

- 1 1 0

1 0 0 1 Step 4 : Take next number (1) from dividend ,

1001(2) =9(10) is greater than divisor i.e. 110(2)



- 1 1 0

1 1 Step 5 : Remainder 11(2) =3(10) is less than



Dividend 100001 33

Divisor 110 6

Quotient 0101 5

Remainder 11 3

Answer is :

Q5 ) 100011 (2) ÷ 10 (2) = ? (2)

Step

1

Step

2

Step

3

Step

4

1 0 0 1

1 0 1 0 0 0 1 1 Step 1 : First 2 Digit of dividend i.e.10 (2) =2(10)

equal to divisor i.e.10(2) =2(10). Thus quotient is

1, after subtraction, reminder is 0 - 1 0

0 0 0 Step 2 : Take next number (0) from dividend

,0(2) =0(10) is less than to divisor i.e.10(2) =2(10).

Thus quotient is 0, get the (0) as it is without

subtraction

- 1 0

0 0 Step 3 : Take next number (0) from dividend,

0(2) =0(10) is less than to divisor i.e.10(2) =2(10).


subtraction

- 1 0

0 1 Step 4 : Take next number (1) from dividend ,

01(2) =1(10) is less than to divisor i.e.10(2) =2(10).


subtraction

- 1 0

0 1 1 Step 5 : Take next number (1) from dividend ,

11(2) =3(10) is greater than divisor i.e. 10(2)


Binary equivalent of is 1(2)

1 0

1 Step 6 : Remainder 1(2) =1(10) is less than

divisor i.e. 10(2) =2(10). Thus stop

111011(2) ÷ 11 (2) = 10011 (2)

100001(2) ÷ 110 (2) = 101 (2)


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Dividend 100011 35

Divisor 10 2

Quotient 1001 17

Remainder 1 1

Answer is :

Binary Division Practice Questions

QUESTIONS ANSWERES

Quotient Remainder

1. 10010011(2) ÷ 1011 (2) = ? (2) 01101 100

2. 10101(2) ÷ 11 (2) = ? (2) 0111 00

3. 1100(2) ÷ 100 (2) = ? (2) 11 00

4. 1111100(2) ÷ 10 (2) = ? (2) 111110 00

5. 11010(2) ÷ 101 (2) = ? (2) 101 01

6. 1011110(2) ÷ 101 (2) = ? (2) 10010 100

7) 11110(2) ÷ 111 (2) = ? (2) 100 10

8) 100000(2) ÷ 10 (2) = ? (2) 10000 00

9) 101101(2) ÷ 11 (2) = ? (2) 1111 00

10) 10110(2) ÷ 1010 (2) = ? (2) 10 10

CONCLUSION Computer is a machine which is based on the principles of

mathematics. Binary number system is primarily

representation of state of each circuit in a computer where

1 means the state is on and 0 means the state is off. When

we type some letters or words, the computer translates

them in numbers as computers can understand only

numbers. The number based conversions are essential in

digital electronics, we have the input in decimal format but it

takes as binary number for the computation. Understanding

arithmetic operation of binary number systems is extremely

useful in many computer-related fields. I encourage you to

become very familiar with binary addition, subtraction,

multiplication and division. I hope you’ve learned a lot from

this article with easy and graphical presentation.

REFERENCES [1]. https://www.researchgate.net/publication/32067764

1_Number_System

[2]. https://www.tutorialspoint.com

[3]. https://swayam.gov.in/nc_details/NPTEL

100011 (2) ÷ 10 (2) = 101 (2)

https://www.researchgate.net/publication/320677641_Number_System

https://www.researchgate.net/publication/320677641_Number_System

https://www.tutorialspoint.com/

https://swayam.gov.in/nc_details/NPTEL

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